ATI RN
Pediatric Genetic Disorders Questions
Question 1 of 5
A person whose karyotype is 45, XX, t(15q;21q). If she mates with a 46, XY normal individual. What is the possibility that they will have a zygote which will develop into Down syndrome? ('t' is translocation):
Correct Answer: A
Rationale: 1/6 (A) is correct. Rationale: Robertsonian t(15;21) carrier (45, XX) produces 6 gamete types: normal 15+21, t(15;21), 15, 21, null 15, null 21. With normal sperm (15+21+Y), only 21+t(15;21) yields 47, +21 (Down). 1/6 gametes result in trisomy 21.
Question 2 of 5
Which banding technique stains heterochromatin (dark) and euchromatin (light)?
Correct Answer: C
Rationale: G-banding (C) stains heterochromatin dark (G-positive) and euchromatin light. Rationale: Giemsa stains AT-rich, gene-poor heterochromatin dark, while GC-rich, gene-active euchromatin stays light, standard for karyotyping.
Question 3 of 5
In which phase of gametogenesis nondisjunction will produce all the cells aneuploidy?
Correct Answer: C
Rationale: Nondisjunction in meiosis I (MI, C) affects all daughter cells. Rationale: MI separates homologs; failure means all gametes get abnormal chromosome numbers (e.g., 24 or 22), unlike MII (sister chromatids, half affected).
Question 4 of 5
A healthy mother is married to a man who has a roberstonian 14,21 balance. What is the possibility to have a baby with down syndrome?
Correct Answer: D
Rationale: 1/6 ( missing in OCR) is correct. Rationale: 45, XY, t(14;21) produces 6 gametes: normal, t(14;21), 14, 21, null 14, null 21. With 46, XX, only 21+t(14;21) yields 47, +21 (Down), 1/6 chance.
Question 5 of 5
Baby with large placenta, with or without microcephaly:
Correct Answer: A
Rationale: Diandric (A, paternal triploidy) fits. Rationale: 2 paternal + 1 maternal sets (e.g., dispermy) cause large placenta, variable fetal size.